University Physics I · Oscillations · 14.6
Simple and Physical Pendulums
A clock pendulum keeps time because every swing takes the same time as the last — nearly. Widen the arc and the promise quietly breaks. Here is where the small-angle period comes from, and how far you can trust it.
Build the model
Connect the measurement to the mechanism.
A pendulum is a rotational oscillator. Gravity applies a restoring torque τ = −Mgd sin θ about the pivot, and τ = I d²θ/dt² turns that into d²θ/dt² = −(Mgd/I) sin θ. That is not SHM: sin θ is not proportional to θ.
Small angles rescue it — below about 10°, sin θ ≈ θ to within half a percent — and the equation collapses to d²θ/dt² = −ω²θ with ω² = Mgd/I, giving T = 2π√(I/(Mgd)) for any rigid body and T = 2π√(L/g) when the mass sits a distance L from the pivot. The mass cancels either way: gravity supplies the torque in proportion to M and the inertia resists in proportion to M. What was discarded comes back as amplitude dependence, T ≈ T₀(1 + θ₀²/16), so isochronism is a first-order result, not a law.
- Simple definition
- A pendulum is a body swinging under a gravitational restoring torque. For small angles the motion is simple harmonic with T = 2π√(I/(Mgd)), which reduces to T = 2π√(L/g) when all the mass sits a distance L from the pivot.
- Example
- Hang a small bob on a 1.00 m string and one full cycle takes 2.01 s. Hang a uniform 1.00 m rod from one end and it takes 1.64 s — same length, but its mass is spread up towards the pivot.
Torque −MgL sin θ divided by inertia ML². The bob's mass cancels before anything is approximated.
Nonlinear — no sinusoidal solution
Length and gravity alone. No mass anywhere, and to first order no amplitude either.
L in m, g in m s⁻², T in s; small angles only
Any rigid body. Both I and the torque scale with M, so the mass cancels again.
I in kg m² about the pivot; d in m, pivot to centre of mass
The simple pendulum that keeps the same time; that point is the centre of oscillation.
T = 2π√(L(eq)/g); shortest period at d = k
The discarded θ³ term returning. Isochronism is first-order, never exact.
θ₀ in radians; +0.19% at 10°, +1.7% at 30°
Constant for a frictionless pivot, and the restoring torque is τ = −dU/dθ.
Small angle: U ≈ ½Mgdθ² — a parabolic well
Gravity as a torque, not a force
Draw the bob's free-body diagram and you get two forces, weight and tension, at an awkward angle to each other. Take torques about the pivot instead and the tension vanishes — its line of action passes through the pivot — leaving one term: τ = −MgL sin θ, with θ measured from the vertical and the minus sign saying the torque always turns the bob back towards the bottom. Newton's second law for rotation, τ = I d²θ/dt² with I = ML², gives ML² d²θ/dt² = −MgL sin θ, or d²θ/dt² = −(g/L) sin θ. Notice what has already happened: M sits on both sides and cancels. Gravity supplies the restoring torque in proportion to mass, and the bob resists it in proportion to mass, so a lead bob and a cork bob on identical strings keep identical time — in vacuum. In air, buoyancy and drag single out the light bob, and the last section takes that up.
The small-angle step, and its price
d²θ/dt² = −(g/L) sin θ is not SHM. The acceleration is proportional to sin θ, not to θ, and the equation has no sinusoidal solution. The series sin θ = θ − θ³/6 + θ⁵/120 − … shows the move: keep the first term only and the equation becomes d²θ/dt² = −(g/L)θ, which is a = −ω²x in angular clothing, with ω = √(g/L) and T = 2π√(L/g). The price is whatever the discarded terms were worth. At θ = 10° (0.1745 rad), sin θ = 0.1736, so writing θ for sin θ overstates the restoring torque by 0.51%; at 30° the overstatement is 4.7%. Keeping the θ³ term is exactly what produces the amplitude correction at the end of this lesson.
Any rigid body: the physical pendulum
Swap the bob on a string for a rigid body hung from a pivot — a swinging leg, a bell, a metre rule on a nail. Nothing in the torque argument changes except the bookkeeping. The weight acts at the centre of mass, a distance d from the pivot, so τ = −Mgd sin θ, and the rotational inertia is I about the pivot rather than ML². The small-angle result is T = 2π√(I/(Mgd)). Get I from the parallel-axis theorem, I = I(cm) + Md². For a uniform rod of length L pivoted at one end, I = ML²/3 and d = L/2, so T = 2π√(2L/(3g)); with L = 1.00 m and g = 9.81 m s⁻² that is 1.64 s, against 2.01 s for a simple pendulum of the same length. Mass still cancels, because for a body of fixed shape I is itself proportional to M.
Equivalent length and the fastest pivot
Write the result as T = 2π√(L(eq)/g) with L(eq) = I/(Md). That is the length of the simple pendulum which would keep the same time, and the point L(eq) below the pivot is the centre of oscillation — strike a bat or a hammer there and the pivot feels no jolt. Substituting I = I(cm) + Md² gives L(eq) = d + k²/d, where k² = I(cm)/M. Both extremes are slow: tiny d means almost no restoring torque, large d means a long pendulum. The minimum sits at d = k, where L(eq) = 2k. A uniform metre rod has k² = L²/12, so its quickest swing comes from a pivot 0.289 m from the centre, giving L(eq) = 0.577 m and T = 1.52 s — shorter than the 1.64 s from the end. Kater's reversible pendulum uses the pair of pivots that share one L(eq) to measure g to five figures.
The energy view: a parabolic well
Take the lowest point as zero. Swinging out to angle θ lifts the centre of mass by d(1 − cos θ), so U = Mgd(1 − cos θ), and with a frictionless pivot E = ½I(dθ/dt)² + Mgd(1 − cos θ) is constant. Expand the cosine: 1 − cos θ = θ²/2 − θ⁴/24 + …, so U ≈ ½Mgdθ². A potential energy quadratic in the displacement is the signature of SHM, and it says the same thing as the linear restoring torque, since τ = −dU/dθ. It also draws the approximation honestly: the true well is slightly wider than that parabola, so a wide swing lingers longer near the ends than SHM predicts. Energy also answers speed questions with no approximation at all — a simple pendulum released from θ₀ passes the bottom at v = √(2gL(1 − cos θ₀)).
Where the small-angle model fails
Keep the θ³ term and the period picks up an amplitude dependence: T ≈ T₀[1 + ¼sin²(θ₀/2)] ≈ T₀(1 + θ₀²/16), with θ₀ in radians. The numbers show why clockmakers cared. At θ₀ = 5° the period runs 0.048% long — 41 s a day. At 10° it is 0.19%, nearly three minutes a day. At 30° it is 1.7%, and at 90° the exact period is 18% above T₀. Near the top the series stops being a correction at all: as θ₀ → 180° the period diverges, because a pendulum balanced exactly upright never comes down. Other assumptions fail more quietly. A bob of real size makes the system a physical pendulum, with T = 2π√((L + k²/L)/g) — slightly slow; a string with mass pulls the centre of mass up instead, and runs slightly fast. Air drag shrinks the amplitude, nudging the period down towards T₀ as the swing dies, and buoyancy trims the effective g.
Change one variable at a time
Make the relationship visible.
Slide the pivot down the rod and watch the period fall to a minimum at d = k = 0.289 m — not at the centre of mass — then climb again; widen the amplitude and the solid curve lifts clear of the dashed small-angle one.
Equivalent length L(eq) = I/(Md)0.577 m
Small-angle period T₀1.524 s
Period at this amplitude1.528 s
Excess over T₀0.27 %
Live interpretationEquivalent length L(eq) = I/(Md): 0.577 m. Small-angle period T₀: 1.524 s. Period at this amplitude: 1.528 s. Excess over T₀: 0.27 %
Catch the common trap
Explain before calculating.
A uniform rod of length L hangs from a frictionless pivot at one end and swings through a small angle. How does its period compare with that of a simple pendulum whose bob hangs a distance L/2 below the same pivot — at the rod's centre of mass?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA simple pendulum has a 0.750 m string and a 250 g bob, and is released through a small arc. Find its period, then say what changes if the bob is swapped for a 500 g one.
- Small angles make the motion SHM with ω = √(g/L), so T = 2π√(L/g). No mass appears in it.
- L/g = 0.750/9.81 = 0.07645 s², and √0.07645 = 0.2765 s.
- T = 2π × 0.2765 = 1.737 s → 1.74 s to three figures.
- Doubling the bob mass changes nothing: gravity supplies the torque in proportion to M and the bob resists in proportion to M, so T stays 1.74 s.
AnswerT = 1.74 s, and the 500 g bob gives the same 1.74 s.
MediumA uniform rod of length 1.20 m and mass 0.80 kg hangs from a frictionless pivot at one end. Find its small-amplitude period and the length of the simple pendulum that keeps the same time.
- Any rigid body swings with T = 2π√(I/(Mgd)), where I is about the pivot and d is the pivot-to-centre-of-mass distance.
- Pivoted at the end, I = ML²/3 and d = L/2, so I/(Mgd) = (ML²/3)/(Mg·L/2) = 2L/(3g) — the 0.80 kg cancels.
- 2L/(3g) = 2.40/29.43 = 0.08155 s²; √0.08155 = 0.2856 s; T = 2π × 0.2856 = 1.794 s → 1.79 s.
- The equivalent length is L(eq) = I/(Md) = 2L/3 = 0.800 m, not the 0.600 m down to the rod's centre of mass.
AnswerT = 1.79 s and L(eq) = 0.800 m; a bob hung at the centre of mass, 0.600 m down, would give 1.55 s — short by the factor √(4/3) = 1.15.
HardA pendulum clock keeps perfect time swinging to an amplitude of 4.0°. After a repair it swings to 9.0°. How much time does it gain or lose in a day?
- The amplitude enters through T ≈ T₀[1 + ¼sin²(θ₀/2)], with θ₀ measured from the vertical — half the total swing.
- At 4.0°: ¼sin²(2.0°) = ¼(0.034899)² = 3.045 × 10⁻⁴, the excess the clock was regulated against.
- At 9.0°: ¼sin²(4.5°) = ¼(0.078459)² = 1.5390 × 10⁻³.
- The period grows by the difference, 1.5390 × 10⁻³ − 3.045 × 10⁻⁴ = 1.2345 × 10⁻³ of itself; a longer period means fewer ticks, so the clock runs slow.
- Over one day: 1.2345 × 10⁻³ × 86 400 s = 106.7 s.
AnswerIt loses about 107 s a day — near enough 1.8 minutes.